What is the center x^2+12x+y^2+12y+71=0 ?

The center x^2+12x+y^2+12y+71=0 is (-6,-6)

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center x^2+12x+y^2+12y+71=0

center

x^{2} + 12 x + y^{2} + 12 y + 71 = 0

(- 6, - 6)

Solution steps

x^{2} + 12 x + y^{2} + 12 y + 71 = 0

Rewrite

x^{2} + 12 x + y^{2} + 12 y + 71 = 0

in the form of the standard circle equation

(x - (- 6))^{2} + (y - (- 6))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- 6, - 6), r = 1

And the center is:

(- 6, - 6)

x^{2} + 6 x + 1 + y^{2} = 4

(x + 4)^{2} + (y + 2)^{2} = 25

(x - 12)^{2} + (y - 7)^{2} = 196

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

(x - 2)^{2} + (y - 5)^{2} = 49

What is the center x^2+12x+y^2+12y+71=0 ?
The center x^2+12x+y^2+12y+71=0 is (-6,-6)